Challenge

You have a sphere 30 cm in diameter.
In it is half a cone, the vertex touching the top of the sphere, and the base, which is 5cm in diameter reaches half way. The cone is sliced vertically.
Now, 4 dots on 4 random positions on the sphere are generated and then connected forming a tetrahedron inside the sphere.

What is the probability that that tetrahedron will contain a part of the cone?

Winners get a free (picture of a) Porsche!
Good luck.

Other urls found in this thread:

github.com/Jojendersie/Epsilon-Intersection
youtube.com/watch?v=OkmNXy7er84
earthobservatory.nasa.gov/Features/OceanCarbon/
mathworld.wolfram.com/SpherePointPicking.html
twitter.com/AnonBabble

How? What? I get the question but dude. Fuck off

I'd say... like 6?

Stop asking brainlet questions on sci. Nobody is gonna fall for solving thus brainlet b8

>can't solve the problem
>tells OP he is a brainlet

Stop projecting, I tried this challenge. It was question 6 on the Putnam Competition that Terrence Tao also competed in. I just beefed it up a bit. It's doable.

Close

Giggled a bit

Everyone here is always raising themselves to Archimedes' level so lets put our money where our mouths are.

>it takes a Terrence Tao level genius to solve this mathematically
>meanwhile a CS student could write a script to brute force the answer
Mathfags btfo

Show your answer*
Checkmate atheists.

I havn't taken a maths course in a decade and I work in banking.

Here's my mathlet crack at it:

Volume of sphere = (4/3)pi*(15^2)
=14137.17 cm^3

Volume of half a cone = (0.5)*((4/3)pi(5*(30/3))
=785.5 cm^3

I don't know how to incorporate the tetrahedron, so lets say the probability is just 785.5/14137.17 = 5.5% or ~1:17

Write it.

If you convert it into 2d you get some clues, but I doubt anyone here can solve it

This shows the relation of volume of the cone vs the sphere yes, but it does not take into account the shape of the tetrahedron indeed, but you said that. Best crack at it so far, only one too

I'd like to refine my answer to

Why less? The tetrahedron is equally likely to have the base bigger in relation to the rest as the vertex. Or is it..? ;)

>Write it.
The premise isn't even clear, it's obvious that OP modified a problem while being shit a writing problems. What is "half a cone" supposed to be in this problem? I still don't understand if the cone is 2D or 3D. let's start by OP posting a relevant diagram of the problem and not something that makes no sense

My thought is that there is more volume of the sphere that doesn't contain the half cone then there is that does, so the tetrahedron has a greater chance of not containing it than it does.

Again, mathlet.

You're right, it's a 3d cone, sliced in half vertically.
It looks like this. And the rest should be clear.

>I still don't understand if the cone is 2D or 3D
Dude, a 2d cone is a triangle. Are you wondering whether the sphere is 3D also? That would be a circle

So the banker is the best mathematician here today

No this is right. And your maths would be absolutely correct if it was a point in the circle, and not a tetrahedron. But you covered that

>Write it.
Not him, but I got half of the problem done. It's around 36% with a whole cone, I was about to use a box to split the cone into two parts and get the half cone intersections but I have to go. Here's the code if anyone wants to clean it up, it should be trivial to finish, if not I'll do it when I come back. The collapsed snippet is just assigning each vector coordinate a random number from the distribution, collapsed to fit more interesting code in the screenshot
I'm using the Epsilon Intersection library ( github.com/Jojendersie/Epsilon-Intersection ) to check for cone-triangle intersections for each triangle of the tetrahedron.

1/8.

Here's the solution. I bet that's where you got th idea from, huh?
youtube.com/watch?v=OkmNXy7er84

By the way, that guy's a good math channel. Can recommend.

Either it intersects the cone, or it doesn't. Probability is 0.5

This is impossible, if it does have an answer you fucked up the question, on a sphere there is no canonical way to pick a random point, especialy since the location of the cone breaks the symmetry of the system. If you give a probability distribution then it's solvable, for example pick the 2 angles with uniform distributions with the point of the cone located at 0,0.

Does anyone have any papers that explore the mechanisms for climate change? like:

earthobservatory.nasa.gov/Features/OceanCarbon/

I want to make up my own mind about this matter, because the usual debate for climate change are just random buzz words.

>on a sphere there is no canonical way to pick a random point
mathworld.wolfram.com/SpherePointPicking.html

Just because wolfram implements it doesn't make it canonical.

Try reading the page before engaging in damage control, role-player.

Sure

My answer is really close to 65%. Am I that much of a brainlet?